Problem: $ \left(\dfrac{64}{27}\right)^{-\frac{1}{3}}$
Solution: $= \left(\dfrac{27}{64}\right)^{\frac{1}{3}}$ Figure out what goes in the blank: $\Big(? \Big)^{3}=\dfrac{27}{64}$ Figure out what goes in the blank: $\Big({\dfrac{3}{4}}\Big)^{3}=\dfrac{27}{64}$ So $\left(\dfrac{64}{27}\right)^{-\frac{1}{3}}=\left(\dfrac{27}{64}\right)^{\frac{1}{3}}=\dfrac{3}{4}$